Fortunately nobody does so in mathematics.

[WM]

Here are all finite initial segments of |N and all their finite initial unions:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

All unions are finite. But the terms of the sequence contain all natural numbers and also all finite initial unions of finite initial segments.

If someone claimed, in mathematics. that |N is somewhat larger, then he'd have to supply evidence in form of an additional something missing everywhere but not in |N. He would fail. Fortunately nobody does so in mathematics.

Regards, WM

[j4n bur53]

Well you started with |N ("initial segments of |N"), and
then you say there is no |N ("that |N is somewhat larger,
would fail").

Doesn't make any sense.

[j4n bur53]

You need to first prove that |N exists,
before you can draw initial segements out of it.

What if |N was only {1,2}, then you even don't
have {1,2,3}. Or if |N is only {1..10^604} where
10^604 is AP brain fartos constant,

then you dont have an open ..., only this here:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

{1} U {1, 2} U {1, 2, 3} .. U {1,..,10^604} = {1,..,10^604}

j4n bur53 schrieb:

[Jew Lover]

On Sunday, May 12, 2019 at 4:18:11 PM UTC-4, WM wrote:
> Here are all finite initial segments of |N and all their finite initial unions:
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> ...
>
> All unions are finite. But the terms of the sequence contain all natural numbers and also all finite initial unions of finite initial segments.

This is classic Euler S = Lim S.

S = {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ... n}
|N = Lim S

or |N = Lim {n -> oo} S for the stupid morons whose inference producing circuitry is dysfunctional.

>
> If someone claimed, in mathematics. that |N is somewhat larger, then he'd have to supply evidence in form of an additional something missing everywhere but not in |N. He would fail. Fortunately nobody does so in mathematics.

I beg your pardon, but they do claim this in mainstream mathematics.

pi = 3.14159...

The limit (whatever the hell it is, because it is not a rational number) hovers mystically above the end of sequence somewhere at infinity (over the rainbow? Chuckle).

Same thing for any other incommensurable magnitude (NOT irrational number because there is no such thing).

>
> Regards, WM

[jvr]

Mücke, you pathetic fool, can you really not learn to understand how
'every', 'each' and 'all' are used in mathematics?

[jvr]

On Sunday, May 12, 2019 at 10:41:55 PM UTC+2, j4n bur53 wrote:
> Well you started with |N ("initial segments of |N"), and
> then you say there is no |N ("that |N is somewhat larger,
> would fail").
>
> Doesn't make any sense.
>

When did anything in Mückenhausen make any sense?

[sergIo]

On 5/13/2019 9:29 AM, Jew Lover wrote:
> On Sunday, May 12, 2019 at 4:18:11 PM UTC-4, WM wrote:
>> Here are all finite initial segments of |N and all their finite initial unions:
>>
>> {1} = {1}
>> {1} U {1, 2} = {1, 2}
>> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
>> ...
>>
>> All unions are finite. But the terms of the sequence contain all natural numbers and also all finite initial unions of finite initial segments.
>
> This is classic Euler S = Lim S.

already to bullshit.

[WM]

Am Sonntag, 12. Mai 2019 22:41:55 UTC+2 schrieb j4n bur53:
> Well you started with |N ("initial segments of |N"), and
> then you say there is no |N

No. I prove that there is no |N that is larger than all FISONs. Since |N cannot be less than any FISON, the result is: |N is a FISON but not a fixed one. it is finite at any point in time, but it grows indefinitely and without bound.

> Doesn't make any sense.

It does not make any sense to believe that |N is larger than all FISONs but not larger than their union. The union is not a magic act. It cannot collect what before was absent.

Regards, WM

[j4n bur53]

But you must already assume an |N that is large
than all FISONs, otherwise you cannot make your list.

[j4n bur53]

If you don't have always:

FISON =< |N

Then you will have some k, such that:

FISON_k = |N

And then your |N was just a FISON and not |N.
|N is not a natural number. Neither omega,
omega is also not a natural number.

[Me]

On Monday, May 13, 2019 at 4:30:03 PM UTC+2, Jew Lover wrote:

> S = {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ... n}

Look, you fucking idiot, since the right side of your equation has a free variable, namely "n", the LHS should have that variable too. In other words, your DEFINITION should read:

S_n := {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ... n}

Why oh why are you dumb like shit, But Sex Lover?

> |N = Lim S

Nope. But

lim_(n->oo) S_n = IN .

> or |N = Lim {n -> oo} S

Again, or |N = lim_{n -> oo} S_n

Hint: Now we may DEFINE

S := lim_{n -> oo} S_n .

Got that, you FUCKING ASSHOLE FULL OF SHIT?

[Jew Lover]

It does not matter how you write it, you moron. All that matters is that you understand. I don't need to cross every t and dot every i so that imbeciles like you feel good about themselves.

Good to know that you finally understand S = Lim S is a bunch of Eulerian nonsense.

[WM]

> > It does not make any sense to believe that |N is larger than all FISONs but not larger than their union. The union is not a magic act. It cannot collect what before was absent.
> >

Am Montag, 13. Mai 2019 22:17:36 UTC+2 schrieb j4n bur53:

> But you must already assume an |N that is large
> than all FISONs, otherwise you cannot make your list.

No, there is no |N. There are only the free standing FISONs.

If an actuallyinfinite |N is assumed, then it is larger than all FISONs. Then |N contains for every FISON and infinite complement. Then |N is much more than the union of all FISONs. But that is disputed too.

A really stupid statement reads: Every FISON has an infinite complement in |N. But the union of all FISONs has no complement in |N. Without severe brain-damage this would not bepossible.

Regards, WM

[WM]

Am Montag, 13. Mai 2019 22:19:12 UTC+2 schrieb j4n bur53:
> If you don't have always:
>
> FISON =< |N
>
> Then you will have some k, such that:
>
> FISON_k = |N

The other way round. Count the FISONs. |N is always the last one.

Regards, WM

[Me]

On Tuesday, May 14, 2019 at 2:30:56 PM UTC+2, Jew Lover wrote:
> On Monday, May 13, 2019 at 7:40:56 PM UTC-4, Me wrote:
> > On Monday, May 13, 2019 at 4:30:03 PM UTC+2, Jew Lover wrote:
> > >
> > > S = {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ... n}
> > >
> > Look, you fucking idiot, since the right side of your equation has a
> > free variable, namely "n", the LHS should have that variable too. In
> > other words, your DEFINITION should read:
> >
> > S_n := {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ... n}

Hint: "S_n" and "S" are different symbols.

> > Now we may DEFINE
> >
> > S := lim_{n -> oo} S_n .
> >

> > Got that [...]?

Hint: "S_n" and "S" are different symbols, denoting different things/objects.

> It does not matter how you write it

Of course it does, idiot.

So at this stage we may claim

[Dan Christensen]

On Sunday, May 12, 2019 at 4:18:11 PM UTC-4, WM wrote:

> Here are all finite initial segments of |N and all their finite initial unions:
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> ...
>

Not this shit again! Are you really going to claim once again for 259th time that, from ZFC, you can derive:

N = U({F1, F2, F3, ...}\{F1, F2, F3, ...} where Fn = {1, 2, ... n}?

I really hope you have learned your lesson, Mucke. Haven't you wasted enough years of your life on this nonsense?


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Dan Christensen]

More absurd quotes from Wolfgang Muckenheim (aka WM or Mucke):

“In my system, two different numbers can have the same value.”
-- sci.math, 2014/10/16

“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20

“1/9 has no decimal representation.”
-- sci.math, 2015/09/22

"0.999... is not 1."
-- sci.logic 2015/11/25

“Axioms are rubbish!”
-- sci.math, 2014/11/19

“Formal definitions have lead to worthless crap like undefinable numbers.”
-- sci.math 2017/02/05

“No set is countable, not even |N.”
-- sci.logic, 2015/08/05

“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05


Slipping ever more deeply into madness...

“There is no actually infinite set |N.”
-- sci.math, 2015/10/26

“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26

“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28

“Everything is in the list of everything and therefore everything belongs to a not uncountable set.” (Huh???)
-- sci.math, 2015/11/30

"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09

“The set of numbers will get empty after all have numbers been used.”
-- sci.math, 2016/08/24

“I need no set theory.”
-- sci.math, 2016/09/01

A special word of caution to students: Do not attempt to use WM's “system” (MuckeMath) in any course work in any high school, college or university on the planet. You will fail miserably. MuckeMath is certainly no shortcut to success in mathematics.

Using WM's “axioms” for the natural numbers, he cannot even prove that 1=/=2. His goofy system is truly a dead-end.


Dan
Download my DC Proof 2.0 software at http://www.dcproof.com

[Ben Bacarisse]

Dan Christensen <Dan_Chr...@sympatico.ca> writes:

> A special word of caution to students: Do not attempt to use WM's
> “system” (MuckeMath) in any course work in any high school, college or
> university on the planet.

Unless you are taking his history course "Geschichte des Unendlichen" at
the Hochschule Augsberg. There you must take care to copy out the right
bit from the class notes into your exam papers or you will not get top
marks.

--
Ben.

[Dan Christensen]

No, "S = Lim S" was your handy-work, Troll Boy. With it, you defaced a copy of the image of the text of Euler's popular article on convergence in big red letters. You continue to this day to insist Euler wrote it! Bizarre beyond belief.

In fact, the notation for limits was not invented until several decades after Euler's death. Didn't count on that, did you, Troll Boy? What a moron.


Dan

[Zelos Malum]

That is some piss poor reasoning. My god WM, how mentally retarded are you?

Just because all your examples are finite does that not mean that |N must be finite and contain something mysterious that none of those finite ones didn't.

That implication does not exist. Only an idiot with the mentality of a child would even think so.

[Zelos Malum]

>This is classic Euler S = Lim S.

Again, none says that.

>or |N = Lim {n -> oo} S for the stupid morons whose inference producing circuitry is dysfunctional.

Something is missing in your notation, can you guess what? Probably not!

>The limit (whatever the hell it is, because it is not a rational number)

ANd it doesn't need to be a rational number.

>hovers mystically above the end of sequence somewhere at infinity (over the rainbow? Chuckle).

Nope. If you understood things you'd not say this.
3.1415...
is notation for a real number, it is not hovering anywhere, it is a real number and we recognise it as the number pi. One can argue that the ... leaves up to some ambiguity and it does but it is salvagable in most instances if the person is not a retard.

However (3,3.1,3.14,3.141,3.1415,...) is a sequence and it MIGHT have a limit, as this is just the sequence with finitely many digits of the decimal expansion of pi, it has the limit of pi in real numbers.

In the cauchy sequence construction of real numbers, WHICH YOU STILL FAILED SHOWING ANYTHIGN WRONG WITH INTERNALLY, it is in the equivalence class of pi, but in this case it is important to note that it does not have any limit as it deals with rational numbers and pi is not rational so in rationals, the sequence has no limit.

But then again you are too stupid to understand the differens ebtween something having a limit in its domain and completing the domain such that there is a natural injection from the origina domain to the new one and in the new one it has a limit.

>Same thing for any other incommensurable magnitude (NOT irrational number because there is no such thing).

Sure is
|R/Q, is the set of irrational numbers.

You do not dictate what is and isn't a number.

[bassam king karzeddin]

On Wednesday, May 15, 2019 at 8:24:39 AM UTC+3, Zelos Malum wrote:
> >This is classic Euler S = Lim S.
>
> Again, none says that.
>
> >or |N = Lim {n -> oo} S for the stupid morons whose inference producing circuitry is dysfunctional.
>
> Something is missing in your notation, can you guess what? Probably not!
>
> >The limit (whatever the hell it is, because it is not a rational number)
>
> ANd it doesn't need to be a rational number.
>
> >hovers mystically above the end of sequence somewhere at infinity (over the rainbow? Chuckle).
>
> Nope. If you understood things you'd not say this.
> 3.1415...
> is notation for a real number, it is not hovering anywhere, it is a real number and

> we recognise it as the number pi.

Are you so dumb up to this limit Zelos Malum? No wonder!

And shamelessly you say "we" instead of only "I", more of wonders!

But if you mean by "we" are only those well-known stubborn Trolls like (Dan, j4n bur 5*s, Python, Bill, S ergo or few more in addition to **YOU**), then you are not blamed in your wrong "belief" about pi being any real existing number

Otherwise get us **today** any decent professional mathematicians with true identity names (if at all existing) who still believe stupidly in such real numbers? wonder!

Almost everybody had already understood this fact "but silently" except your group of big Morons

That is why nobody would ever dare to say it openly any more like before

And for you so specially, you are not blamed at all to say anything in this regard since you were already proven "idiot" beyond imaginations, and for sure

> One can argue that the ... leaves up to some ambiguity and it does but it is salvagable in most instances if the person is not a retard.
>
> However (3,3.1,3.14,3.141,3.1415,...) is a sequence and it MIGHT have a limit, as this is just the sequence with finitely many digits of the decimal expansion of pi, it has the limit of pi in real numbers.
>

You don't understand the real number meaning to be illegible to talk about for sure

> In the cauchy sequence construction of real numbers, WHICH YOU STILL FAILED SHOWING ANYTHIGN WRONG WITH INTERNALLY, it is in the equivalence class of pi, but in this case it is important to note that it does not have any limit as it deals with rational numbers and pi is not rational so in rationals, the sequence has no limit.

You still don't understand what are you talking about for sure, so better keep silent in this issue since it isn't for you defiantly

>
> But then again you are too stupid to understand the differens ebtween something having a limit in its domain and completing the domain such that there is a natural injection from the origina domain to the new one and in the new one it has a limit.
>

It is indeed too funny how a real BIG STUPID see others as too stupid too, No wonder!

Areal victim Zelos of wrong global teaching in mathematics FOR SURE

> >Same thing for any other incommensurable magnitude (NOT irrational number because there is no such thing).
>
> Sure is
> |R/Q, is the set of irrational numbers.
>
> You do not dictate what is and isn't a number.

It is more than clear that your own real numbers were a matter of dictating and nothing else For sure,

Zelos, you are certainly a very lost and hopeless case despite many remedy courses just before you, FOR SURE
BKK

[Jew Lover]

As I said many times, my article describes in detail what S = Lim S means and no doubt you've read it, but here you are still driveling about what it means.

Therefore, you are a moron who can't be helped.

https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel

[Jew Lover]

On Wednesday, May 15, 2019 at 1:18:12 AM UTC-4, Zelos Malum wrote:
> Den söndag 12 maj 2019 kl. 22:18:11 UTC+2 skrev WM:
> > Here are all finite initial segments of |N and all their finite initial unions:
> >
> > {1} = {1}
> > {1} U {1, 2} = {1, 2}
> > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > ...
> >
> > All unions are finite. But the terms of the sequence contain all natural numbers and also all finite initial unions of finite initial segments.
> >
> > If someone claimed, in mathematics. that |N is somewhat larger, then he'd have to supply evidence in form of an additional something missing everywhere but not in |N. He would fail. Fortunately nobody does so in mathematics.
> >
> > Regards, WM
>
> That is some piss poor reasoning. My god WM, how mentally retarded are you?
>
> Just because all your examples are finite does that not mean that |N must be finite

Correct. S = Lim S shows that |N does not exist.

> and contain something mysterious that none of those finite ones didn't.

>

> Only an idiot with the mentality of a child would even think so.

Of course you should know because you are a child.

[Jew Lover]

On Wednesday, May 15, 2019 at 1:24:39 AM UTC-4, Zelos Malum wrote:
> >This is classic Euler S = Lim S.
>
> Again, none says that.

It says EXACTLY that and explained in detail what it means many times.

https://www.linkedin.com/pulse/eulers-worst-definition-lim-john-gabriel

You're just too stupid and stubborn.

[Zelos Malum]

>Are you so dumb up to this limit Zelos Malum? No wonder!

The dumb one is you, you get NOTHING right in mathematics.

>And shamelessly you say "we" instead of only "I", more of wonders!

We, as in everyone not a fucking idiot.

>But if you mean by "we" are only those well-known stubborn Trolls like (Dan, j4n bur 5*s, Python, Bill, S ergo or few more in addition to **YOU**), then you are not blamed in your wrong "belief" about pi being any real existing number

I mean people with half a brain, which excludes you. And pi is provably existing.

>Otherwise get us **today** any decent professional mathematicians with true identity names (if at all existing) who still believe stupidly in such real numbers? wonder!

You can go to any mathematician at any university.

>Almost everybody had already understood this fact "but silently" except your group of big Morons

Yeah you are delusional if you think so. The reason they don't botherwith you is becuase you have NOTHING to come with. YOu ahve not presented any proofs, no arguements, NOTHING of value. You do not even know what "Exists" means in formal mathematics.

>That is why nobody would ever dare to say it openly any more like before

They do you moron, everywhere.

>And for you so specially, you are not blamed at all to say anything in this regard since you were already proven "idiot" beyond imaginations, and for sure

The idiot here is you, again I can cite sources on EVERYTHIGN I say, can you? No you cannot. You cannot even cite it for ANYTHING of yours.

>You still don't understand what are you talking about for sure, so better keep silent in this issue since it isn't for you defiantly

I understand it perfectly thank you. Want me to cite sources?

>Areal victim Zelos of wrong global teaching in mathematics FOR SURE

Typical conspiratard thinking there is a global conspiracy, rather than facing the fact that they are an idiot and wrong.

>Zelos, you are certainly a very lost and hopeless case despite many remedy courses just before you, FOR SURE

Again, I got an education in mathematics and knows this shit, what do you have? Oh yeah you are just a stupid engineer!

[Zelos Malum]

>As I said many times, my article describes in detail what S = Lim S means and no doubt you've read it, but here you are still driveling about what it means.

And you are wrong there as we have pointed out.

Want me to do it again? Unlike you I will state the actual places where you go off the rail.

>Correct. S = Lim S shows that |N does not exist.

There is no such implication, first of all because S = Lim S, is wrong and non-sense, secondly |N's existence is axiomatic.

>It says EXACTLY that and explained in detail what it means many times.

No it does not, Rudin and many others write

S=Lim S_n, which is a shorthand for S=lim (S_n)_{n e |N}

but I doubt you'd understand the nuasances of such.

[Me]

On Thursday, May 16, 2019 at 3:03:14 AM UTC+2, Jew Lover wrote:

> As I said many times, my article describes in detail what S = Lim S means ...

Yes, but you "description" is nonsense.

If (S_n) is a sequence, we may define/write

S := lim_(n->oo) S_n .

If c is a constant real number we MIGHT write

c = lim_(n->oo) c .

[Me]

On Thursday, May 16, 2019 at 3:05:05 AM UTC+2, Jew Lover wrote:

> S = Lim S shows that

you are a moron.

[Jew Lover]

On Thursday, May 16, 2019 at 1:33:44 AM UTC-4, Zelos Malum wrote:
> >As I said many times, my article describes in detail what S = Lim S means and no doubt you've read it, but here you are still driveling about what it means.
>
> And you are wrong there as we have pointed out.

You are wrong as I have pointed out.

First of all, you can't write Lim S_n = S because Lim S_n does not get a value assigned to it. Rather it returns a value. But because you know Euler's S = Lim S is bogus, you have attempted to prolong your deception by trying to reverse the assignment. You fail yet again.

Partial sums cannot be assigned the series - that is very circular, but to you circular reasoning works because what does a moron like you know...

[WM]

Am Mittwoch, 15. Mai 2019 07:18:12 UTC+2 schrieb Zelos Malum:
> Den söndag 12 maj 2019 kl. 22:18:11 UTC+2 skrev WM:
> > Here are all finite initial segments of |N and all their finite initial unions:
> >
> > {1} = {1}
> > {1} U {1, 2} = {1, 2}
> > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > ...
> >
> > All unions are finite. But the terms of the sequence contain all natural numbers and also all finite initial unions of finite initial segments.
> >
> > If someone claimed, in mathematics. that |N is somewhat larger, then he'd have to supply evidence in form of an additional something missing everywhere but not in |N. He would fail. Fortunately nobody does so in mathematics.
> >

> Just because all your examples are finite does that not mean that |N must be finite

The examples are all finite and they contain all natural numbers. An infinite set is larger.

Try to answer using mathematical arguments, not credo in absurdum: What makes |N larger than all unions of all FISONs?

Regards, WM

[Zelos Malum]

>You are wrong as I have pointed out.

Nope, it is you who are wrong, always ahve been, always will be.

>because Lim S_n does not get a value assigned to it

It has a value assigned to it, if it exist because Lim is a partial function in most cases.

the value it is assigned, r, is the value such that AxEnAm(m>n => |r-S_n|<x)

That is the assigned value, if it exists. For a cauchy sequence it always does.

>But because you know Euler's S = Lim S

Again, none writes that because it is non-sense.

>you have attempted to prolong your deception by trying to reverse the assignment. You fail yet again.

No one is doing any deception, if you actually took the time to read and understand, you'd see you are the one misunderstanding.

>Partial sums cannot be assigned the series - that is very circular, but to you circular reasoning works because what does a moron like you know...

You do not know what circular means.

THere is nothing circular in

"Given the following notation
Sum_{i e |N} a_i, of a sequence (a_i)_{i e |N}, we construct the sequence (Sum_{j=0}^i a_j)_{i e |N} and assign it the value of Lim (Sum_{j=0}^i a_j)_{i e |N}, which may be written as Lim a_i as well"

There is no circularity, there is notational jumps and such there for our convinience.

[Zelos Malum]

>The examples are all finite and they contain all natural numbers. An infinite set is larger.

Noen of those contains all natural numbers.

>Try to answer using mathematical arguments, not credo in absurdum: What makes |N larger than all unions of all FISONs?

The union of all FISONS (a set that is in bijection with |N, mind you) is still |N

[WM]

Am Freitag, 17. Mai 2019 07:21:27 UTC+2 schrieb Zelos Malum:
> >The examples are all finite and they contain all natural numbers. An infinite set is larger.
>
> Noen of those contains all natural numbers.

Then some natural number should be missing from all. What natural number is missing from all FISONs? None. Therefore all natural numbers are listed here:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

>

> >Try to answer using mathematical arguments, not credo in absurdum: What makes |N larger than all unions of all FISONs?
>
> The union of all FISONS (a set that is in bijection with |N, mind you) is still |N

The union of all FISONs is in the above sequence. None is missing. (You cannot find a missing one. Pure contrary belief is not accepted.) But the sequence does not contain any term that is larger than all FISONs. |N however is believed to be larger than all FISONs.

Regards, WM

[jvr]

[...]

>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> ...

>

> The union of all FISONs is in the above sequence.

So for some n we have {1,2, ... , n} = |N.
Please let us know for which n this is true in Mückenhausen.

[WM]

Am Freitag, 17. Mai 2019 18:55:26 UTC+2 schrieb jvr:
> [...]
> >
> > {1} = {1}
> > {1} U {1, 2} = {1, 2}
> > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > ...
>
> >
> > The union of all FISONs is in the above sequence.
>
> So for some n we have {1,2, ... , n} = |N.

There is no maximum of the infinite sequence.

> Please let us know for which n this is true in Mückenhausen.

Do you claim that there are more FISONs and more gapless unions of FISONs existing in mathematics than are exsisting in the sequence?

Please let me know which are missing.

Regards, WM

[Zelos Malum]

>Then some natural number should be missing from all

Incorrect, just because each FISON does not contain all natural numbers does that not mean that each natural number is in at least 1 FISON.

[WM]

Am Montag, 20. Mai 2019 07:39:03 UTC+2 schrieb Zelos Malum:
> >Then some natural number should be missing from all
>
> Incorrect, just because each FISON does not contain all natural numbers does that not mean that each natural number is in at least 1 FISON.

Each natural nuber is in infinitely many FISONs, therefore it is also in at least 1 FISON.

The sequence

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

contains all natural numbers (as elements of FISONs) and all FISONs (at the right-hand side) and all gapless unions of FISONs (at the left-hand side). There is no infinite union of FISONs and no infinite set of natural numbers.

If there are more natural numbers than are contained in all FISONs, then they must be somewhere else. If there are not more then the set N is not actually infinite.

Regards, WM

[Me]

On Monday, May 20, 2019 at 5:26:57 PM UTC+2, WM wrote:

> The sequence
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> ...

WHICH sequence, you moron?

The sequence of equations

"{1} = {1}", "{1} U {1, 2} = {1, 2}", "{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}" ...

or the sequence

{1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, ...

or the sequence

{1}, {1, 2}, {1, 2, 3}, ...

(which actually is identical with the former)?

> contains all natural numbers (as elements of FISONs) and all FISONs (at the
> right-hand side) and all gapless unions of FISONs (at the left-hand side).

This is just some idiotic mumbo-jumbo. Next time try to formulate a mathematical claim, itiot.

> There is no infinite union of FISONs and no infinite set of natural numbers.

This may be true in your WMmath, but in set theory BOTH sets exist (actually, they are identical).

[William]

WN is caught in the standard crank loop.

C: A is true. A implies B. B is true.

NC: No A is true. A does not imply B. B is False

C: Oh you think B is false. Then you must think A is false. I will prove A.

In this case A is "Each natural number is in some FISON" and B is "|N is not actually infinite". Thus WM spends a large number of electrons showing that every natural number is in some FISON.

--
Willam Hughes

[Zelos Malum]

>Each natural nuber is in infinitely many FISONs, therefore it is also in at least 1 FISON.

Correct, but no FISON contains all natural numbers.

>contains all natural numbers (as elements of FISONs) and all FISONs (at the right-hand side)

Correct, even a blind chicken can have luck it seems!

>There is no infinite union of FISONs and no infinite set of natural numbers.

Only in the given sequence.

>If there are more natural numbers than are contained in all FISONs, then they must be somewhere

There isn't, but the fact that your sequence of FISONs do not contain |N does not in anyway imply that |N does not exist.

[WM]

Am Dienstag, 21. Mai 2019 07:43:33 UTC+2 schrieb Zelos Malum:
> >Each natural nuber is in infinitely many FISONs, therefore it is also in at least 1 FISON.
>
> Correct, but no FISON contains all natural numbers.

Name a natural number that is not in a single FISON together with all other natural numbers that can be named. (A counter example would need two natunumbers thqt are not in any FISON.)

>
> >contains all natural numbers (as elements of FISONs) and all FISONs (at the right-hand side)
>

> >There is no infinite union of FISONs and no infinite set of natural numbers.
>
> Only in the given sequence.

The given sequence contains all gapless unions of FISONs. More FISONs do not exist. (Otherwise you could show a counter example.)

>
> >If there are more natural numbers than are contained in all FISONs, then they must be somewhere
>
> There isn't, but the fact that your sequence of FISONs do not contain |N does not in anyway imply that |N does not exist.

Your |N is a farce. And if not, it contains mainly dark numbers that cannot be named. All natnumbers that can be named are in one FISON.

Regards, WM

[WM]

Am Dienstag, 21. Mai 2019 02:52:21 UTC+2 schrieb William:

WH is caught in the standard crank loop.

The facts are clear.

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

All unions are finite. But the terms of the sequence contain, as elements of FISONs, all natural numbers that can be specied, all FIS (on the right-hand side), and also all gapless unions of these FIS (on the left-hand side).

But WH does not like these facts and calls them names.

If you deny that all natural numbers are in a single FISON, then name a natural number that is not in a single FISON together with all other natural numbers that can be named. A counter example would need two natunumbers that are not in any FISON.

Regards, WM

[WM]

Am Montag, 20. Mai 2019 21:42:17 UTC+2 schrieb Me:
> On Monday, May 20, 2019 at 5:26:57 PM UTC+2, WM wrote:
>
> > The sequence
> >
> > {1} = {1}
> > {1} U {1, 2} = {1, 2}
> > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > ...
>
> WHICH sequence,

There is only one, written in two different ways.

> > There is no infinite union of FISONs and no infinite set of natural numbers.
>
> This may be true in your WMmath,

Above we see that all unions of FISONs are finite. More FISONs are not available.

> but in set theory BOTH sets exist (actually, they are identical).

They are not in the above sequence. |N is *claimed* to be something larger than all unions of FISONs of the above sequence. Not yet proven.

Regards, WM

[William]

Here is the Wolkenmuekenhim proof that all natural numbers are in one
FISON

The sequence

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

is a potentially infinite sequence. That means it always has a last
FISON. The Last FISON clearly contains all natural numbers in the FISONS
in the list. Every natural number is in some FISON. Thus the natural numbers are contained in the last FISON. (The last FISON is a Wolkenmuekenheim special, something that changes. The natural numbers are not contained in a fixed FISON.)

WM is also fond of the false statement: If there is no FISON that contains all natural numbers then there must be two natural numbers that are not in a single FISON. This does not follow. Any set of two natural numbers, indeed any set of natural numbers with a largest element, is contained within a single FISON. That does not mean that any set of natural numbers without a largest element is contained in a single FISON. [Again there is no problem in Wolkenmuekenheim where the natural numbers are potentially infinite, and thus any existing set of natural numbers has a largest element]

The FISON

{1,2,3,...,Largest existing Natural}

is a (changing) FISON that contains all existing natural numbers if and only if tha Largest existing Natural always exists.

--
William Hughes

[horand....@gmail.com]

This seems to be a stronger requirement than the notion underlying ultrafinitism, namely that the set N does not exists. Maybe we should call this philosophy "transfinitism". (SCNR)

[WM]

Am Dienstag, 21. Mai 2019 16:41:24 UTC+2 schrieb William:
>
> The sequence
>
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> ...
>
> is a potentially infinite sequence.

That is not what I prove. My proof is this: If the set |N is actually infinite, that is: larger than all FISONs, then this |N contains dark numbers.

> That means it always has a last
> FISON. The Last FISON clearly contains all natural numbers in the FISONS
> in the list. Every natural number is in some FISON. Thus the natural numbers are contained in the last FISON. (The last FISON is a Wolkenmuekenheim special, something that changes.

That is not speciality of mine but known to every good mathematician like Cantor, Hilbert, ..., Simpson:

"In spite of significant difference between the notions of the potential and actual infinite, where the former is a variable finite magnitude, growing above all limits," [Cantor, p. 374]

"In analysis we have to deal only with the infinitely small and the infinitely large as a limit-notion, as something becoming, emerging, produced, i.e., as we put it, with the potential infinite." [D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p. 167]

"A potential infinity is a quantity which is finite but indefinitely large. For instance, when we enumerate the natural numbers as 0, 1, 2, ..., n, n+1, ..., the enumeration is finite at any point in time, but it grows indefinitely and without bound." [S.G. Simpson: "Potential versus actual infinity: insights from reverse mathematics" (2015)]

> The natural numbers are not contained in a fixed FISON.)

Correct. The maximum F(m) is increasing forever, but never becoming infinite and never more than one FISON.

> WM is also fond of the false statement: If there is no FISON that contains all natural numbers then there must be two natural numbers that are not in a single FISON. This does not follow.

Perhaps not in matheology in the theory of Zero Findable Contradictions. In ordinary logic it follows. What does it mean, that not all natnumbers are contained? That means according to current logic that at least one natnumber is not contained. Since one is contained, we have two numbers that are not in a single FISON.

> Any set of two natural numbers, indeed any set of natural numbers with a largest element, is contained within a single FISON.

All natnumbers that can be specified can be found in FISONs.

> That does not mean that any set of natural numbers without a largest element is contained in a single FISON.

Name a natnumber or a FISON that is not in the infinite sequence:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

All the contents of the sequence, all natnumbers contained in FISONs, all FISONs contained in unions, all that is contained in one single FISON. The right-hand side does not contain your magic unions that allegedly are larger than all unioned elements.

> [Again there is no problem in Wolkenmuekenheim where the natural numbers are potentially infinite, and thus any existing set of natural numbers has a largest element]
>
> The FISON
>
> {1,2,3,...,Largest existing Natural}
>
> is a (changing) FISON that contains all existing natural numbers if and only if tha Largest existing Natural always exists.

Otherwise we have evidence for dark numbers.

Regards, WM

[William]

On Tuesday, May 21, 2019 at 2:14:50 PM UTC-3, WM wrote:
> Am Dienstag, 21. Mai 2019 16:41:24 UTC+2 schrieb William:

<snip>

> > WM is also fond of the false statement: If there is no FISON that contains all natural numbers then there must be two natural numbers that are not in a single FISON. This does not follow.
>
> Perhaps not in matheology in the theory of Zero Findable Contradictions. In ordinary logic it follows. What does it mean, that not all natnumbers are contained? That means according to current logic that at least one natnumber is not contained.

Nope, using ordinary logic, saying that the set of natnumbers are not contained means: (i) there is one natnumber that is not contained *or* (ii) the set of natnumbers has no largest element. (i) is false but (ii) is true.


--
William Hughes

[WM]

Why should this handwaving and unfounded claim be accepted outside of matheology?

There is no largest natnumber. But we can use a variable n running through all natnumbers.

All natnumbers that exist are subject to logic and universal quantification. In particular all are in FISONs.

Then this n will never cease to be in one and the same FISON with all natnumbers.

For natnumbers that cannot be specified other rules may apply.

Regardes, WM

[William]

On Tuesday, May 21, 2019 at 5:23:57 PM UTC-3, WM wrote:
> Am Dienstag, 21. Mai 2019 20:41:08 UTC+2 schrieb William:


> > Nope, using ordinary logic, saying that the set of natnumbers are not contained means: (i) there is one natnumber that is not contained *or* (ii) the set of natnumbers has no largest element. (i) is false but (ii) is true.

> >
>
> Why should this handwaving and unfounded claim be accepted outside of matheology?

It is easy. (i) is obvious. (ii) follows directly from: All FISONS have a largest element. No set without a largest element is contained within a set with largest element.

>
> There is no largest natnumber.

Indeed and there is no largest specifiable natural number.


Outside of Wolkenmuekenheim there is a set that includes every specifiable natural number.

--
William Hughes

[Me]

On Tuesday, May 21, 2019 at 7:14:50 PM UTC+2, WM wrote:

> My proof is this: [...]

You don't have any proofs, idiot.

[Me]

On Tuesday, May 21, 2019 at 4:03:13 PM UTC+2, WM wrote:
> Am Montag, 20. Mai 2019 21:42:17 UTC+2 schrieb Me:
> > On Monday, May 20, 2019 at 5:26:57 PM UTC+2, WM wrote:
> > >
> > > The sequence
> > >
> > > {1} = {1}
> > > {1} U {1, 2} = {1, 2}
> > > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > > ...
> > >

> > WHICH sequence, idiot?

> >
> There is only one, written in two different ways.

Idiotic blather.

Hint: We may consider the sequences

(A_i) with A_i = {1} U ... U {1, ..., n} for all i e IN
and
(B_i) with B_i = {1, ..., n} for all i e IN .

Then (A_i) = (B_i) since A_i = B_i for all i e IN.

But the sequence of equations

"{1} = {1}", "{1} U {1, 2} = {1, 2}", "{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}", ...

is a completely different sequence.

So which sequence are you takling about, idiot?

> we see that all unions of FISONs are finite.

No, "we" don't *see* that, idiot.

But we _know_ that all *finite* unions of FISIONs (i.e. all unions of finitely many FISONs) are finite.

In the same way we _know_ that all *infinite* unions of FISIONs (i.e. all unions of infinitely many FISONs) are infinite.

We _know_ these two facts because we can PROVE them in the context of set theory.

> IN is [...] larger than all [finite] unions of FISONs [...].

Right, since IN is infinite, but all unions of finitely many FISONs are finite.

[Zelos Malum]

>Name a natural number that is not in a single FISON together with all other natural numbers that can be named.

All natural numbers is in some FISON but no FISON has all the natural numbers.

>The given sequence contains all gapless unions of FISONs

Nope, it contains all FINITE gapless unions. Finite being a keyword here.

>Your |N is a farce

Not at all, you being too stupid to understand basic set theory does not invalidate it.

>And if not, it contains mainly dark numbers that cannot be named

Incorrect, it contains all the numbers and all can be named.

>All natnumbers that can be named are in one FISON.

All natural numbers is in a FISON and the union of ALL fisons, infinitely many of them, is |N

[WM]

Am Dienstag, 21. Mai 2019 23:54:11 UTC+2 schrieb William:
> On Tuesday, May 21, 2019 at 5:23:57 PM UTC-3, WM wrote:
> > Am Dienstag, 21. Mai 2019 20:41:08 UTC+2 schrieb William:
>
>
> > > Nope, using ordinary logic, saying that the set of natnumbers are not contained means: (i) there is one natnumber that is not contained *or* (ii) the set of natnumbers has no largest element. (i) is false but (ii) is true.
>
> > >
> >
> > Why should this handwaving and unfounded claim be accepted outside of matheology?
>
>
> It is easy. (i) is obvious. (ii) follows directly from: All FISONS have a largest element. No set without a largest element is contained within a set with largest element.

Here is your error. There is no largest element. Therefore we do not state anything about the largest element. But all existing elements that have a finite specification, are not largest elements and are subject to my proof.

Note that we only consider natnumbers that can be specified and are in FISONs with all such elements.

>
>
> >
> > There is no largest natnumber.
>
> Indeed and there is no largest specifiable natural number.

But each one is in the sequence - and the sequences contains only finite sets.

>
>
> Outside of Wolkenmuekenheim there is a set that includes every specifiable natural number.

Here is the sequence of finite sets. F(1), F(2), F(3), ... It is infinite by never ending. But that does not make it larger than all FISONs. |N however is believed to be larger.

Regards, WM

[WM]

Am Dienstag, 21. Mai 2019 23:54:11 UTC+2 schrieb William:

> No set without a largest element is contained within a set with largest element.

But all natnumbers which can be specified and are in FISONs are in one FISON.

Regards, WM

[WM]

Am Mittwoch, 22. Mai 2019 07:22:58 UTC+2 schrieb Zelos Malum:
> >Name a natural number that is not in a single FISON together with all other natural numbers that can be named.
>
> All natural numbers is in some FISON but no FISON has all the natural numbers.

What natnumbers can be specified that are not in one of the FISONs together?

>
> >The given sequence contains all gapless unions of FISONs
>
> Nope, it contains all FINITE gapless unions. Finite being a keyword here.

If you know more FISONs that can be specified, please specify them

>
> >Your |N is a farce
>

> >And if not, it contains mainly dark numbers that cannot be named
>
> Incorrect, it contains all the numbers and all can be named.

Then name some natnumbers that are not in one FISON together.

>
> >All natnumbers that can be named are in one FISON.
>

> All natural numbers is in a FISON and the union of ALL fisons, infinitely many of them, is |N.

Proof by assertion? Not valid! Please specify some FISONs that are not in a finite union of the following sequence of unions. Fail.

{1}
{1, 2}
{1, 2, 3}
...

Regards, WM

[WM]

Am Mittwoch, 22. Mai 2019 01:10:06 UTC+2 schrieb Me:
> On Tuesday, May 21, 2019 at 4:03:13 PM UTC+2, WM wrote:
> > Am Montag, 20. Mai 2019 21:42:17 UTC+2 schrieb Me:
> > > On Monday, May 20, 2019 at 5:26:57 PM UTC+2, WM wrote:
> > > >
> > > > The sequence
> > > >
> > > > {1} = {1}
> > > > {1} U {1, 2} = {1, 2}
> > > > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > > > ...
> > > >
> > > WHICH sequence, idiot?
> > >
> > There is only one, written in two different ways.
>
> Idiotic blather.

Obvious mathematical fact because

∀n ∈ ℕ: {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, n} = {1, 2, 3, n}

>
>
> > we see that all unions of FISONs are finite.
>
> No, "we" don't *see* that,

Every mathematician would see it.

> But we _know_ that all *finite* unions of FISIONs (i.e. all unions of finitely many FISONs) are finite.

There are no other unions of FISONs as is proved by the left-hand side of the sequence.

>
> In the same way we _know_ that all *infinite* unions of FISIONs (i.e. all unions of infinitely many FISONs) are infinite.

Alas, there are no infinite unions of FINITE initial segments. See the sequence above. Please insert further FISONs if you know them.

>
> We _know_ these two facts because we can PROVE them in the context of set theory.
>
> > IN is [...] larger than all [finite] unions of FISONs [...].
>
> Right, since IN is infinite, but all unions of finitely many FISONs are finite.

Therefore your |N must contain more than can be specified. Or can you specify natnumbers that are not elements of the FISONs of the right-hand side or elements of the FINITE unions of FISONs at the left-hand side?

Regards, WM

[WM]

Not for the fantasies of matheologians. But every sober mind knows that all natnumbers which can be specified and are in FISONs are in one FISON.

Proof: There is no natnumber that can be specified outside of all FISONs. Note: This holds also for the union of all FISONs, namely one term of the sequence

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

Regards, WM

[William]

On Wednesday, May 22, 2019 at 5:02:30 AM UTC-3, WM wrote:
> Am Dienstag, 21. Mai 2019 23:54:11 UTC+2 schrieb William:

<snip>

>
> > Outside of Wolkenmuekenheim there is a set that includes every specifiable natural number.
>
> Here is the sequence

Nope. I said set, not sequence of sets. Outside of Wolkenmuekenheim there is a single set |N = F(1) U F(2) U F(3) ... This set does not have a largest element. Outside of Wolkenmuekenheim, all FISONs are fixed, so all FISONs have a largest element. No set without a largest element is contained in a set with a largest element. So |N is not contained in a FISON.

--
William Hughes

[Me]

On Wednesday, May 22, 2019 at 12:29:08 PM UTC+2, WM wrote:

> [...] all natnumbers [...] are in one FISON.

No, idiot. For each and every natural number n there is a FISON F such that n is in F. But there's no FISON F such that for each and every natural number n n is in F.

True: An e IN EF e FISION: n e F.

False: EF e FISION An e IN: n e F.

Hint:

An e IN: {1, ..., n} e FISON & n e {1, ..., n} and

AF e FISON: max(F)+1 e IN & max(F)+1 !e F ,

dumbo.

[WM]

Am Mittwoch, 22. Mai 2019 13:50:57 UTC+2 schrieb William:
> On Wednesday, May 22, 2019 at 5:02:30 AM UTC-3, WM wrote:

> > Here is the sequence

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

>

> Nope. I said set, not sequence of sets.

All possible sets and all possible unions of FISONs are limited to terms of the above sequence. Otherwise try to determine what could generate an infinite union of FISONs, i.e., a larger union than all possible unions being present in the above sequence on the left-hand side.

> Outside of Wolkenmuekenheim there is a single set |N = F(1) U F(2) U F(3) ... This set does not have a largest element.

Where is it? If it is existing outside of matheological belief too, then it can be determined what makes it larger than all finite unions of FISONs given abobe. Try to show it. Pure statement of believe is insufficient.

> Outside of Wolkenmuekenheim, all FISONs are fixed, so all FISONs have a largest element.

Of course. But there is no last FISON.

> No set without a largest element is contained in a set with a largest element. So |N is not contained in a FISON.

In the course of this discussion, even from this posting of mine you can obtain the simple truth: There is no |N larger than all FISONs of the sequence, and there is no union of FISONs larger than all FISONs. All that is rubbish. But there is no greatest natnumber and no greatest FISON.

Regards, WM

[WM]

Am Mittwoch, 22. Mai 2019 23:02:17 UTC+2 schrieb Me:
> On Wednesday, May 22, 2019 at 12:29:08 PM UTC+2, WM wrote:
>
> > [...] all natnumbers [...] are in one FISON.
>

> No. For each and every natural number n there is a FISON F such that n is in F. But there's no FISON F such that for each and every natural number n n is in F.

Prove it. Show a union of natnumbers that is larger than all unions of FISONs of the following sequence (left-hand side):

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

Note that all these unions are FISONs (right-hand side).

>
> True: An e IN EF e FISION: n e F.
>
> False: EF e FISION An e IN: n e F.
>
> Hint:
>
> An e IN: {1, ..., n} e FISON & n e {1, ..., n} and
>
> AF e FISON: max(F)+1 e IN & max(F)+1 !e F ,
>

But they are all in the next FISONs.

Again, try to comprehend what I say: Show a union of definable natnumbers (not your folly |N - that is mainly consisting of dark numbers) that is not in one and the same FISON.

Regards, WM

[William]

On Thursday, May 23, 2019 at 9:01:45 AM UTC-3, WM wrote:
> Am Mittwoch, 22. Mai 2019 13:50:57 UTC+2 schrieb William:

<snip>

> > Outside of Wolkenmuekenheim there is a single set |N = F(1) U F(2) U F(3) ... This set does not have a largest element.
>

> Where is it? If it is existing outside of matheological belief too, then it can be determined what makes it larger than all finite unions of FISONs given above.


The thing that makes |N larger is the fact that it does not have a largest element.[Note |N is not larger because it contains some element not in the finite union of FISONs]


More detail


A set with no largest element is always larger (not contained in)

a set with largest element.

We are not in Wokenmuekenheim so every element of |N exists. |N does not have a largest element. Each finite union of FISONs has a largest element. |N is larger than any finite union of FISON because it does not have a largest element.


--
William Hughes

[jvr]

[...]

>
> A set with no largest element is always larger (not contained in)
> a set with largest element.
>

That can't be what you meant to say.

[WM]

Am Donnerstag, 23. Mai 2019 17:39:33 UTC+2 schrieb William:
> On Thursday, May 23, 2019 at 9:01:45 AM UTC-3, WM wrote:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

> > Where is it? If it is existing outside of matheological belief too, then it can be determined what makes it larger than all finite unions of FISONs given above.
>
>
> The thing that makes |N larger is the fact that it does not have a largest element.

Sets are defined by elements that they have, not by elements they don't have.

Further the set of all finite unions of FISONs does not have a last element either.

[Note |N is not larger because it contains some element not in the finite union of FISONs]

In order to be larger than every finite union, |N must have at least one element not in any finite union of FISONs.

>
>
> More detail
>
>
> A set with no largest element is always larger (not contained in)
> a set with largest element.

Sets are defined by elements that they have, not by elements they don't have.

>
> every element of |N exists. |N does not have a largest element. Each finite union of FISONs has a largest element.

Irrelevant since the set of all finite unions of FISONs does not have a last element either.

> |N is larger than any finite union of FISON because it does not have a largest element.

|N cannot be larger than all finite unions of FISONs because there is no natnumber outside of all finite unions of FISONs.

Further: Sets are no defined by missing elements but by elements they have. Therefore find an element outside of every finite union of FISONs. Fail. Recognize that actual infinity (fixed number of more elements than all FISONs have) is a chimera.

Regards, WM

[William]

On Thursday, May 23, 2019 at 1:55:30 PM UTC-3, WM wrote:
> Am Donnerstag, 23. Mai 2019 17:39:33 UTC+2 schrieb William:
> >
> >
> > The thing that makes |N larger is the fact that it does not have a largest element.
>
> Sets are defined by elements that they have, not by elements they don't have.

There is a function L:|N -> |N, s.t. for each k in |N, L(k) defined and existing, k<L(k). For each k in |N, k and L(k) are elements that |N has.

>
> Further the set of all finite unions of FISONs does not have a last element either.
>

Your claim is not that !N is contained in the set of all finite unions of
FISONS, but in a single element of the set of all finite unions of FISONS.
Every element of this set has a largest member.

>> [Note |N is not larger because it contains some element not in the finite union of FISONs]
>
> In order to be larger than every finite union, |N must have at least one element not in any finite union of FISONs.

*or* there must be a function L:|N -> |N, for each k in |N, L(k) defined and existing, k<L(k)

--
William Hughes

[jvr]

[...]

>
> [Note |N is not larger because it contains some element not in the finite union of FISONs]
>
> In order to be larger than every finite union, |N must have at least one element not in any finite union of FISONs.

Please, Mücke, I know you are not very bright but are you really so dim
that you believe this simplistic fallacy?

But if you don't believe it, why are you presenting it as though it were
a valid argument?

I guess the answer is: Yes, poor Mücke really is that stupid.

[WM]

Am Donnerstag, 23. Mai 2019 20:17:33 UTC+2 schrieb William:
> On Thursday, May 23, 2019 at 1:55:30 PM UTC-3, WM wrote:
> > Am Donnerstag, 23. Mai 2019 17:39:33 UTC+2 schrieb William:
> > >
> > >
> > > The thing that makes |N larger is the fact that it does not have a largest element.
> >
> > Sets are defined by elements that they have, not by elements they don't have.
>
> There is a function L:|N -> |N,

My proof shows that there is no such |N. So your function is fantasy, not prving anything.

> s.t. for each k in |N, L(k) defined and existing, k<L(k). For each k in |N, k and L(k) are elements that |N has.

All defined natural numbers together belong to a FISON. Proof: All definable numbers are elements in finite unions of FISONs

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

That proves that the set consisting of definable numbers is not actually infinite.

>
> >
> > Further the set of all finite unions of FISONs does not have a last element either.
> >
>
> Your claim is not that !N is contained in the set of all finite unions of
> FISONS, but in a single element of the set of all finite unions of FISONS.

Of course. There are only finite unions. More is not possible. That is not a claim but easy to see from the sequence.

> Every element of this set has a largest member.

Of course. And every natural number that can be defined belongs together with all other natural numbers that can be defined to one and the same FISON. To contradict this, you have to define a natural number that does not belong together with all others to any FISON. You cannot. Therefore you are wrong. But you are a crank who is not able to understand his mistakes.

> > In order to be larger than every finite union, |N must have at least one element not in any finite union of FISONs.
>
> *or* there must be a function L:|N -> |N, for each k in |N, L(k) defined and existing, k<L(k)

That function is not existing. And if existing, then the required natunumbers cannot be defined. All definable nunmbers are in one and the same FISON. To contradict this, you need no mysterious functions on not existing sets but at least one definable natnumber that is not in the same FISOn as 1.

Note, in logic we contradict a universal claim not by functions but by showing the existence of a counterexample. Only cranks do not know about this habit of serious logic.

Regards, WM

[WM]

Am Donnerstag, 23. Mai 2019 20:55:46 UTC+2 schrieb jvr:


> > In order to be larger than every finite union, |N must have at least one element not in any finite union of FISONs.
>

> this simplistic fallacy?

In logic there is a simple rule, and you need not be bright to know it: If there is a universal claim to be disproved, then a simple counterargument has to be shown.

Example: All definable natnumbers belong to one and the same FISON. Show a counterexample or recognize your mistake. Note: We are talking about definable natnumbers, not about matheologial nonsense.

Regards, WM

[jvr]

Right, Mücke. You can imagine how grateful we all are to be able to learn
all about Muckmeatics and Logic, real genuine Aristotelian Logic, from such
a great expert.
Particularly the subtleties that have ruined traditional mathematics, the
confusion of each and every, the mistaken belief that Lim Card != Card Lim,
and the paint job on the famous muckmeatical Xmas tree, your brilliant
McDuck example...
Let us all praise great men!

[William]

On Thursday, May 23, 2019 at 4:41:33 PM UTC-3, WM wrote:

<snip>

> All [the set of] defined natural numbers together belong to a FISON. Proof: All [each one of the] definable numbers are elements in finite unions of FISONs

Standard ambiguous use of all.

--
William Hughes

[Me]

On Thursday, May 23, 2019 at 9:48:54 PM UTC+2, WM wrote:

> In logic there is a simple rule, and you need not be bright to know it:

> If there is a universal claim to be disproved, then <bla>
>
> Example: All definable natnumbers belong to one and the same FISON. (*)

Holy shit! No, Mückenheim, this is NOT a universal claim, but an existential claim. (Your quantor dyslxia strikes again.)

Hint: (*) might be formalized the following way:

Ef e F: An e DN: n e F .

There is a FISON f such that all "definable" natural numbers are in f.

[Zelos Malum]

>Here is the sequence of finite sets. F(1), F(2), F(3), ... It is infinite by never ending. But that does not make it larger than all FISONs. |N however is believed to be larger.

|N has greater cardinality than any FISON, however it is in a bijection (aka same cardinality) as the set of all FISONs.

No one argue against that.

>What natnumbers can be specified that are not in one of the FISONs together?

As stated, any natural number is in some FISON, but no FISON has all natural numbers.

>Then name some natnumbers that are not in one FISON together.

I do not need to do such because I already stated that given any natural number, there is some FISON containing it.

However that does nto mean that you can give me a FISON containing all natural numbers.

>Proof by assertion? Not valid! Please specify some FISONs that are not in a finite union of the following sequence of unions. Fail.

Nope, if you want a proof I can give it, it is extremely trivial. But then again easy mathematics is not your forté now is it? or really any mathematics.

Given any finite union of FISONs, it is in your list.

But your list does not contain the union of all FISONs

>Every mathematician would see it.

They won't beacuse most things you say are blantantly false when you go by strict logic and not gut feeling intuition.

>There are no other unions of FISONs as is proved by the left-hand side of the sequence.

Incorrect, you can take many unions of INFINITELY many fisons, yours only contain FINITELY many fisons.

>Alas, there are no infinite unions of FINITE initial segments. See the sequence above. Please insert further FISONs if you know them.

Only in your list aren't there more. But I can make an infinite union of FISONs if I want to.

>Therefore your |N must contain more than can be specified. Or can you specify natnumbers that are not elements of the FISONs of the right-hand side or elements of the FINITE unions of FISONs at the left-hand side?

Non-sequitor, just because no finite union of FISONs is |N does that not mean that |N contains anything but what you can find in the FISONs.

This is another example of you failing mathematics.

[WM]

Am Freitag, 24. Mai 2019 07:29:17 UTC+2 schrieb Zelos Malum:


> >What natnumbers can be specified that are not in one of the FISONs together?
>
> As stated, any natural number is in some FISON, but no FISON has all natural numbers.

For every collection of definable natnumbers there is a FISON containing this collection. Whether this collection is a set is irrelevant.

If |N is larger, then it contains undefinable dark numbers.

> >Then name some natnumbers that are not in one FISON together.
>
> I do not need to do such because I already stated that given any natural number, there is some FISON containing it.

Not any, but all! That is the point.

>
> However that does nto mean that you can give me a FISON containing all natural numbers.

I can give you the theroem that all natnubers that you or anybody else can define belong to one and the same FISON.

>
> >Proof by assertion? Not valid! Please specify some FISONs that are not in a finite union of the following sequence of unions. Fail.

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

>

> Nope, if you want a proof I can give it,

I want an example.

> But your list does not contain the union of all FISONs

Give an example of a FISON not contained by the finite unions.

>
>
> >There are no other unions of FISONs as is proved by the left-hand side of the sequence.
>
> Incorrect, you can take many unions of INFINITELY many fisons, yours only contain FINITELY many fisons.

The infinite union should be larger than all finite unions. So there must be more FISONs than in any finite union. Name one!

>
> >Alas, there are no infinite unions of FINITE initial segments. See the sequence above. Please insert further FISONs if you know them.
>
> Only in your list aren't there more. But I can make an infinite union of FISONs if I want to.

Do it. But all FISONs are already in my finite unions. The infinite union should be larger than all finite unions. So there must be more FISONs than in any finite union. Name one!

>
> >Therefore your |N must contain more than can be specified. Or can you specify natnumbers that are not elements of the FISONs of the right-hand side or elements of the FINITE unions of FISONs at the left-hand side?
>
> Non-sequitor, just because no finite union of FISONs is |N does that not mean that |N contains anything but what you can find in the FISONs.
>

If |N contains nothing more than all finite unions of FISONs, what makes it larger than all finite unions of FISONs? Sets are determined by their elements.

Regards, WM

[WM]

Am Donnerstag, 23. Mai 2019 22:56:15 UTC+2 schrieb William:
> On Thursday, May 23, 2019 at 4:41:33 PM UTC-3, WM wrote:
>
>
> > All [the set of] defined natural numbers together belong to a FISON.

All definable natural numbers together belong to one and the same FISON. Whether they make up a set is ambiguous and irrelevant.

> Proof: All [each one of the] definable numbers are elements in finite unions of FISONs
>
> Standard ambiguous use of all.

Not at all! Your approach to lie fails misarably. Fact is:

All definable natnumbers together are elements of one and the same FISON. Proof: Try to find a counterexample. Fail.

Regards, WM

[WM]

Am Donnerstag, 23. Mai 2019 23:14:01 UTC+2 schrieb Me:
> On Thursday, May 23, 2019 at 9:48:54 PM UTC+2, WM wrote:
>
> > In logic there is a simple rule, and you need not be bright to know it:
> > If there is a universal claim to be disproved, then <bla>
> >
> > Example: All definable natnumbers belong to one and the same FISON. (*)
>
> Holy shit! No, Mückenheim, this is NOT a universal claim, but an existential claim.

My claim that nobody can specify a counterexample is proved by the fact that nobody can find a counterexample.

>
> Hint: (*) might be formalized the following way:
>

For every collection of definable natnumbers there is a FISON containing this collection. Whether this collection is a set is irrelevant.

Regards, WM

[Python]

Crank Professor Wolfgang Mueckenheim from Hochschule Augsburg wrote:
...

> For every collection of definable natnumbers there is a FISON containing this collection. Whether this collection is a set is irrelevant.

O Dear! Crank Professor Wolfgang Mueckenheim from Hochschule Augsburg,
usually you are the one pretending to make a point in a alleged proof
that "x being a member of a set" can be used as an argument (like
"being a member of a finite set"), now - out of you sophistry - you
happen to hate that being a set for a collection so far you call it
"irrelevant"??

How come, Crank Professor Wolfgang Mueckenheim from Hochscule Augsburg?
You, the only silly mind on Earth pretending that "being a member of a
finite set" is /per se/ a "relevant" property?

Crank Professor Wolfgang Mueckenheim from Hochschule Augsburg, your
sophistry is going worse and worse.

[Jew Lover]

On Friday, May 24, 2019 at 7:43:06 PM UTC-4, Crank Jean Pierre Messger (aka Python) driveled:

When will you ever learn you moron? "Sophistry" is not a well-formed word.

Sophistry: the clever use of arguments that seem true but are really false, in order to deceive people.

https://dictionary.cambridge.org/dictionary/english/sophistry

Here's the gist you idiot: arguments that are false will never seem true to a rational mind. Ah, I forget ... I am talking to an idiot.

[WM]

Am Samstag, 25. Mai 2019 01:43:06 UTC+2 schrieb Python:

> usually you are the one pretending to make a point in a alleged proof
> that "x being a member of a set" can be used as an argument (like
> "being a member of a finite set"),

Here is a sequence of finite sets:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

that does contain all definable numbers but does neither contain an actually infinite union of FISONs nor an actually infinite set of FISONs.

Therefore you have to find something that make an actially infinite union or to agrre that there is no finished infinity.

Regards, WM

[Me]

On Friday, May 24, 2019 at 11:12:50 PM UTC+2, WM wrote:

> For every collection of [...] natnumbers there is a FISON containing this
> collection.

Nope. IN is a "collection" of natnumbers. But there is no FISON F suc that IN c F.

[Zelos Malum]

>For every collection of definable natnumbers there is a FISON containing this collection. Whether this collection is a set is irrelevant.

Incorrect, for any FINITE collection of definable natural numbers there is a FISON.

You assume definable implies finite, which is faulty.

>If |N is larger, then it contains undefinable dark numbers.

Incorrect, ser previous.

>Not any, but all! That is the point.

Nope, given any natural number, tehre is a FISON containing it, but given all natural numbers, there is no FISON containing them.

>I can give you the theroem that all natnubers that you or anybody else can define belong to one and the same FISON.

Except you cannot.

>I want an example.

I can give a proof, do you want it or nto? A proof may contain an example, it may not. The relevans is that I can logically prove it to be true.

>Give an example of a FISON not contained by the finite unions.

I never argued that, I said that your list only contains finite union, not all unions. All unions include the infinite cases.

>The infinite union should be larger than all finite unions. So there must be more FISONs than in any finite union. Name one!

The infinite union is larger than any FISON and any finite union of FISONs. There are no other FISONs than those provided.

>Do it. But all FISONs are already in my finite unions. The infinite union should be larger than all finite unions. So there must be more FISONs than in any finite union. Name one!

Nope, you only got FINITE unions of FISONS, not the infinite ones.

Are you seriously this stupid that you do not understand that your list only contains finite ones?

>If |N contains nothing more than all finite unions of FISONs, what makes it larger than all finite unions of FISONs? Sets are determined by their elements.

The fact it is not finite. For any finite union of FISONs you have a finite number of elements. Every endomorphism that is injective is also surjective.

when you take the union of all you get a set where it no longer holds true.

[WM]

Am Montag, 27. Mai 2019 07:33:59 UTC+2 schrieb Zelos Malum:
> >For every collection of definable natnumbers there is a FISON containing this collection. Whether this collection is a set is irrelevant.
>
> Incorrect, for any FINITE collection of definable natural numbers there is a FISON.
>
> You assume definable implies finite, which is faulty.

I prove that all definable FISONs yield finite unions of FISONs.

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...
>

> Nope, given any natural number, tehre is a FISON containing it, but given all natural numbers, there is no FISON containing them.

Natnumbers from FISONs are all in the above unions.

>
> >Give an example of a FISON not contained by the finite unions.
>
> I never argued that, I said that your list only contains finite union, not all unions. All unions include the infinite cases.

Then say what the finite unions makes infinite.

>
> >The infinite union should be larger than all finite unions. So there must be more FISONs than in any finite union. Name one!
>
> The infinite union is larger than any FISON and any finite union of FISONs. There are no other FISONs than those provided.

They fail to yield an infinite union.

>
> >Do it. But all FISONs are already in my finite unions. The infinite union should be larger than all finite unions. So there must be more FISONs than in any finite union. Name one!
>
> Nope, you only got FINITE unions of FISONS, not the infinite ones.

I got all unions.

>
> Are you seriously this stupid that you do not understand that your list only contains finite ones?

That is clear. Show what you will use to increase the unions!

Regards, WM

[WM]

Then try to find a FISON or what you may believe that makes the finite unions of the following sequence infinite:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

Or recognize that your |N is not a collection of natnumbers from FISONs.

Regards, WM

[Me]

On Monday, May 27, 2019 at 6:10:53 PM UTC+2, WM wrote:
> Am Montag, 27. Mai 2019 07:33:59 UTC+2 schrieb Zelos Malum:
> >
> > The infinite union is larger than any FISON and any finite union of FISONs.
> > There are no other FISONs than those provided.
> >
> They fail to yield an infinite union.

No, they dont, you imbecile. Actually, *YOU* failed to CONSIDER an infinite union.

Look, idiot, here's such an "infinite union":

U{{1,...,n} : n e IN} .

Again, the set of all fisions is

AF := {{1,...,n} : n e IN} .

Then the union of this (infinite) set, i.e. the "infinite union", is

U AF = IN .

> > you [on the other hand --me] only got FINITE unions of FISONS, not the

> > infinite ones.
> >
> I got all unions.

Nope, dumbo. For example, you didn't consider the (infinite) union

U{{1,...,n} : n e IN}

you only considered the (infinitely many) FINITE unions:

U{{1}}
U{{1}, {1,2}}
U{{1}, {1,2}, {1,2,3}
:

> > Are you seriously this stupid that you do not understand that your list
> > only contains finite ones?

Seems so.

[Zelos Malum]

>I prove that all definable FISONs yield finite unions of FISONs.

No you didn't. You proved that any finite amount of FISONs is a finite union (SHOCKER!)

There is nothing in it that prohibits an infinite union.

>Natnumbers from FISONs are all in the above unions.

but none of those FISONs contains all natural numbers.

if you pick a FISON and claim it has all natural numbers, I can find a natural number that fison does not have.

>Then say what the finite unions makes infinite.

Why? That is not what we argued, again your list only contains finite unions, not all unions.

>They fail to yield an infinite union.

Not at all
U{F_0,F_1,F_2,...}=|N

That is a union of infinitely many FISONs and it is an infinite set.

>I got all unions.

You didn't, no where in your list is U{F_0,F_1,F_2,...}

>That is clear. Show what you will use to increase the unions!

the axioms of ZFC and I can construct {F_0,F_1,F_2,...} and actually even P(|N) many kinds of unions, albeit almost all there will still just yield |N

[WM]

Am Montag, 27. Mai 2019 20:14:53 UTC+2 schrieb Me:
> On Monday, May 27, 2019 at 6:10:53 PM UTC+2, WM wrote:
> > Am Montag, 27. Mai 2019 07:33:59 UTC+2 schrieb Zelos Malum:
> > >
> > > The infinite union is larger than any FISON and any finite union of FISONs.
> > > There are no other FISONs than those provided.
> > >
> > They fail to yield an infinite union.
>

> No, they dont. Actually, *YOU* failed to CONSIDER an infinite union.

Does the sequence of natural numbers 1, 2, 3, ... contain an infinite number? No.

Does the sequence of unions of FISONs

{1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, ... (*)

contain an infinite union? No.

Proof, the FISONs and the unions of FISONs are in bijection with the natural numbers.

The sequence of unions of FISONs contains all gapless unions of FISONs. All are finite. Unions with gaps contain less FISONs than unions without gaps. An infinite union however contains more FISONs than every finite union.

> here's such an "infinite union":
>
> U{{1,...,n} : n e IN} .

Does it exist? No. An infinite union must contain more FISONs than every finite union. But sequence (*) exhausts the set of all FISONs and shows that all FISONs are insufficient to produce an infinite union.

Othrwise there must be more than in all terms of sequence (*).

> Again, the set of all fisions is
>
> AF := {{1,...,n} : n e IN} .

Yes. All these FISONs are unioned in (*) and are not producing an infinite union.

>
> Then the union of this (infinite) set

Here you ar mistaken. Either you can find more FISONs than are unioned in (*) or your claim is void and is an unoproven statment which is known to carry little weight in mathmatics.

> > I got all unions.
>
> Nope,

Do you understand that the sequenc (*) has only finite unions?
Do you agree that all FISONs are unioned in (*)?

> you only considered the (infinitely many) FINITE unions:
>
> U{{1}}
> U{{1}, {1,2}}
> U{{1}, {1,2}, {1,2,3}
> :

How could these finite unions become infinite? By pure claim? By some magic? By insulting the advocate of the superior argument until he will withdraw?

>
> > > Are you seriously this stupid that you do not understand that your list
> > > only contains finite ones?
>
> Seems so.

Seems so.

Show some substance, i.e., FISONs, that transform the infinite sequence (*) of finite unions into an infinite union.

Regards, WM

[WM]

Am Dienstag, 28. Mai 2019 08:10:17 UTC+2 schrieb Zelos Malum:
> >I prove that all definable FISONs yield finite unions of FISONs.
>
> No you didn't. You proved that any finite amount of FISONs is a finite union (SHOCKER!)

The shock comes if you se that no further FISONs ar available to increase the finit terms of the sequence

{1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, ... (*)

>
> There is nothing in it that prohibits an infinite union.

All FISONs are exhaustd by finite unions in (*). There is nothing remaining for a union larger than any finite union.

>
> >Natnumbers from FISONs are all in the above unions.
>
> but none of those FISONs contains all natural numbers.

The terms of the sequence (*) are strictly increasing. They do not lose FISONs. All FISONs that are contained are in one term.

>
> if you pick a FISON and claim it has all natural numbers, I can find a natural number that fison does not have.

I do not pick a FISON? Why should I? I considr all possible (gapless) unions of FISONs.

>
> >Then say what the finite unions makes infinite.
>
> Why?

My sequence (*) contains all unions. If you claim that not all unions are ther, thn shows what FISON is missing.

> That is not what we argued, again your list only contains finite unions, not all unions.

That is what you argue: There are more FISONs than are unioned in (*).

>
> >They fail to yield an infinite union.
>
> Not at all
> U{F_0,F_1,F_2,...}=|N

F_0, F_1, F_2, ... are all unioned in (*). By bijection with |N this is shown. No trm in infinite. Therefore you claim that there is more. You must show it because I claim that there is not more. Your actual infinity is unfounded belief.

>

> >That is clear. Show what you will use to increase the unions!
>
> the axioms of ZFC and I can construct {F_0,F_1,F_2,...}

Do you? Do you start to construct the sequence {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, ... ? Where do you surpass it?

> and actually even P(|N) many kinds of unions,

Start with the gapless unions of the sequence (*) and mark the point where you surpass it.

Regards, WM

[Python]

Herr Crank Wolfgang Muckenheim, from Hochschule Augsburg, wrote:
> Am Dienstag, 28. Mai 2019 08:10:17 UTC+2 schrieb Zelos Malum:
>>> I prove that all definable FISONs yield finite unions of FISONs.
>>
>> No you didn't. You proved that any finite amount of FISONs is a finite union (SHOCKER!)
>
> The shock comes if you se that no further FISONs ar available to increase the finit terms of the sequence
>
> {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, ... (*)
>>
>> There is nothing in it that prohibits an infinite union.
>
> All FISONs are exhaustd by finite unions in (*). There is nothing remaining for a union larger than any finite union.

Quantifier dislexia again, Herr Crank Wolfgang Muckenheim, from
Hochschule Augsburg?

[WM]

Am Dienstag, 28. Mai 2019 18:28:20 UTC+2 schrieb Python:
> Herr Crank Wolfgang Muckenheim, from Hochschule Augsburg, wrote:
> > Am Dienstag, 28. Mai 2019 08:10:17 UTC+2 schrieb Zelos Malum:
> >>> I prove that all definable FISONs yield finite unions of FISONs.
> >>
> >> No you didn't. You proved that any finite amount of FISONs is a finite union (SHOCKER!)
> >
> > The shock comes if you se that no further FISONs ar available to increase the finit terms of the sequence
> >
> > {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, ... (*)
> >>
> >> There is nothing in it that prohibits an infinite union.
> >

> > All FISONs are exhausted by finite unions in (*). There is nothing remaining for a union larger than any finite union.
>
> Quantifier dislexia

No, simple logic. If the infinite union is larger than all gapless finite unions, then the infinite union must contain more than all gapless finite unions.

Regards, WM

[William]

Each gapless finite union (do not use "all". it confuses you).


--
William Hughes

[WM]

The sequence contains *all* gapless unions of definable FISONs. There are not more. If you don't understand that, then try to define more.

Regards, WM

[konyberg]

If the infinite union is larger than all finite unions, then it contains all finite unions.
If there is a finite union you haven't thought about? It will be there! :) By definition.
KON

[Python]

Le 29/05/2019 à 17:52, William a écrit :

Crank Wolfgang Mueckenheim, from Hochschule Augsburg, won't do so, this
is one of the pillars of his sophistry.

[WM]

Am Mittwoch, 29. Mai 2019 20:39:31 UTC+2 schrieb konyberg:
> onsdag 29. mai 2019 17.40.12 UTC+2 skrev WM følgende:


> If the infinite union is larger than all finite unions, then it contains all finite unions.

They are contained already in the sequence

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
...

No infinite union in sight though.

> If there is a finite union you haven't thought about?

The sequence contains all gapless finite unions. Unions with gaps are irrelevant because they are smaller.

> It will be there! :) By definition.

All gapless unions are there. No actually infinite union is there. And it can't be constructed from definable FISONs. Otherwise someone would have done so.

Regards, WM